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There is a famous integrals (sinz)/z dz with limit from -α(infinity) to α.
Now in complex analysis is done in way where the take sinz = Im(eiz ) and then calculate the residue. And it come out to be π My question is that can't we calculate residue by the method where we find the coefficient of 1/z in doing expansion of the whole function. Here expansion of sinz = z - (z3)/(3!) ...... When multiplied by 1/z (in question) then there is no coefficient of 1/z. Which should make the residue=0. What am I missing here, please help me out.
Here are the two methods I am discussing about IMG-20240713-232541.jpg
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